Problem: Let $y=\ln(\sin(x))$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{\sin(x)}$ (Choice B) B $\dfrac{\cos(x)}{\sin(x)}$ (Choice C) C $\dfrac{1}{\cos(x)}$ (Choice D) D $\ln(\cos(x))$
$\ln(\sin(x))$ is a composition of two, more basic, functions: $\sin(x)$ and $\ln(x)$. In other words, suppose $u(x)=\sin(x)$ and $v(x)=\ln(x)$, then $\ln(\sin(x))=v\bigl(u(x)\bigr)$, or $(v\circ u)(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the chain rule : $\begin{aligned} \dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&=\dfrac{dv}{du}\cdot\dfrac{du}{dx} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x) \end{aligned}$ Finding $v'\Bigl(u(x)\Bigr)$ $v(x)=\ln(x)$, and therefore $v'(x)=\dfrac{1}{x}$. Now we plug $u(x)=\sin(x)$ into $v'$ : $\begin{aligned} v'\Bigl(u(x)\Bigr)&=v'\Bigl(\sin(x)\Bigr) \\\\ &={\dfrac{1}{\sin(x)}} \end{aligned}$ Finding $u'(x)$ $u(x)=\sin(x)$, and therefore $u'(x)={\cos(x)}$. Putting things together $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\ln(\sin(x))\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=\sin(x)\text{, }v(x)=\ln(x)} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{The chain rule}} \\\\ &={\dfrac{1}{\sin(x)}}\cdot{\cos(x)} \\\\ &=\dfrac{\cos(x)}{\sin(x)} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\cos(x)}{\sin(x)}$.